Subsidies and pricing strategies in a vehicle scrappage program with strategic consumers

Abstract

We consider the problem of a government that wishes to promote replacing old cars with new ones via a vehicle scrappage program. Since these programs increase consumer’s willingness to pay for a new car, manufacturers (or dealers) could respond strategically by raising their prices. In a two-period game between a government and a manufacturer, we find equilibrium prices and subsidy levels. Our results demonstrate that price levels are increasing over time and are higher than in the benchmark case where no subsidy is offered. Furthermore, if consumers act strategically, then the equilibrium price levels will be higher than in the scenario where they behave myopically.

Appendix: Proofs

1.1 Proof of Proposition 1

We start by solving for the manufacturer to obtain its reaction to the subsidy announced by the government. Recall that the manufacturer’s profit is given by

$$\begin{aligned} \Pi =\Pi _{1}+\Pi _{2}=p_{1}(1-2p_{1}+\left( \theta -\gamma \right) s)+p_{2}\left( 2p_{1}+\gamma s-p_{2}\right) . \end{aligned}$$

(14)

First, we consider the second-period manufacturer’s profit. Taking the derivative of $\Pi _{2}$ with respect to $p_{2}$, we get

$$\begin{aligned} \frac{\partial \Pi _{2}}{\partial p_{2}}=0\Leftrightarrow p_{2}=\frac{ 2p_{1}+\gamma s}{2}. \end{aligned}$$

Substituting for $p_{2}$ in $\Pi$ yields

$$\begin{aligned} \Pi =p_{1}(1-2p_{1}+\left( \theta -\gamma \right) s)+\left( \frac{ 2p_{1}+\gamma s}{2}\right) ^{2}. \end{aligned}$$

Differentiating with respect to $p_{1}$ gives

$$\begin{aligned} \Pi ^{\prime }=1-2p_{1}+\theta s=0\Leftrightarrow p_{1}=\frac{1+\theta s}{2}. \end{aligned}$$

Substituting in $p_{2}$, we obtain the manufacturer’s reaction functions, that is,

$$\begin{aligned} p_{1}= & {}\; \dfrac{1}{2}\left( \theta s+1\right) , \nonumber \\ p_{2}= & {}\; \dfrac{1}{2}\left( \gamma s+\theta s+1\right) . \end{aligned}$$

(15)

To solve the government’s problem, we write down the Lagrangian

$$\begin{aligned} {\mathcal {L}}\left( s,\mu \right) = s.d_{1}^{\ge \eta }+s.d_{2}+\mu (\Gamma -d_{1}^{\ge \eta }-d_{1}^{\eta -1}-d_{2}), \end{aligned}$$

where $\mu$ is the Lagrange multiplier appended to the target constraint. Developing the above equation and inserting for $p_{1}$ and $p_{2}$ from (15), we get

$$\begin{aligned} {\mathcal {L}}\left( s,\mu \right) =\Gamma \mu -\frac{1}{2}\mu +s\beta +\frac{1}{ 2}s^{2}\gamma -\frac{1}{2}s\theta \mu +\frac{1}{2}s\gamma \mu . \end{aligned}$$

Assuming an interior solution, the first-order optimality conditions are

$$\begin{aligned}{} & {} \frac{\partial {\mathcal {L}}}{\partial s}=0\Leftrightarrow 2\beta +2s\gamma +\mu \left( \gamma -\theta \right) =0, \\{} & {} \frac{\partial {\mathcal {L}}}{\partial \mu }=0\Leftrightarrow 2\Gamma -1-s\left( \theta -\gamma \right) =0. \end{aligned}$$

Solving, we obtain

$$\begin{aligned}{} & {} s=\frac{2\Gamma -1}{\theta -\gamma }, \\{} & {} \mu =\frac{2\left( \left( 2\Gamma -1\right) \gamma +\beta \left( \theta -\gamma \right) \right) }{\left( \theta -\gamma \right) ^{2}}. \end{aligned}$$

The values of $p_{1}$ and $p_{2}$ can be found by inserting s into their formulas.

1.2 Proof of Proposition 2

As in Proposition 1, we solve for the manufacturer and get the following reaction function:

$$\begin{aligned} p=\dfrac{1}{2}\left( \theta s+1\right) . \end{aligned}$$

To solve the the government’s problem, we introduce the Lagrangian

$$\begin{aligned} {\mathcal {L}}=s(s\theta +\beta -p)+s(\gamma s+p)+\mu (-s\theta +\Gamma +p-1), \end{aligned}$$

where $\mu$ is the Lagrange multiplier. Substituting for p and solving, we obtain the following unique subsidy and value of the Lagrange multiplier:

$$\begin{aligned} {\tilde{s}}= & {}\; \frac{2\Gamma -1}{\theta }>0,\\ {\tilde{p}}= & {}\; \Gamma >0, \\ \mu= & {}\; \frac{2(4\Gamma \gamma +4\Gamma \theta +\beta \theta -2\gamma -2\theta )}{\theta ^{2}}. \end{aligned}$$

Substituting in the price, manufacturer’s profit and government’s cost, we obtain the ${\tilde{p}}$, ${\tilde{\Pi }}$ and ${\tilde{C}}$ in the statement of the Proposition. We note that the equilibrium is indeed interior, and the demands are positive:

$$\begin{aligned} d_{1}= & {}\; \frac{\gamma \left( 1-2\Gamma \right) }{\theta }>0, \\ d_{2}= & {}\; \frac{1}{\theta }\left( \theta \Gamma +\gamma \left( 2\Gamma -1\right) \right) >0. \end{aligned}$$

1.3 Proof of Proposition 4

Same as proposition 1, we solve for the manufacturer and we get the following price functions:

$$\begin{aligned} p_{1}= & {}\; \dfrac{1}{2}\theta s_{2}+\dfrac{1}{2}\\ p_{2}= & {}\; \dfrac{1}{2}\gamma s_{2}+\theta s_{2}-\dfrac{1}{2}\theta s_{1}+ \dfrac{1}{2}. \end{aligned}$$

The Lagrangian function of the government is given by:

$$\begin{aligned} {\mathcal {L}}=&\,s_{1}\left( \theta s_{1}+\beta -\dfrac{1}{2}\theta s_{2}-\dfrac{1 }{2}\right) +s_{2}\left( \dfrac{1}{2}\gamma s_{2}-\dfrac{1}{2}\theta s_{1}+\theta s_{2}+\dfrac{1}{2}\right) \\ &\quad +\mu (-s_{2}\theta +\Gamma +p_{2}-1), \end{aligned}$$

which results in the following solution:

$$\begin{aligned} \check{s}_{1}= & {}\; \frac{2\theta \left( 2\Gamma -1\right) +\gamma \left( 2\beta -1\right) )}{2\theta (\theta -2\gamma )}>0,\\ \check{s}_{2}= & {}\; \frac{2\beta +8\Gamma -5}{2\left( \theta -2\gamma \right) } >0,\\ \mu= & {}\; -\dfrac{1}{2}\frac{8\Gamma \gamma +12\Gamma \theta +4\beta \theta -6\gamma -7\theta }{\theta (2\gamma -\theta )}, \end{aligned}$$

which leads to the following prices:

$$\begin{aligned} \check{p}_{1}= & {}\; \frac{(2\beta +8\Gamma -3)\theta -4\gamma }{4\left( \theta -2\gamma \right) }=\frac{2\left( \left( \beta +\Gamma \right) \theta -2\gamma \right) +(6\Gamma -3)\theta }{4\left( \theta -2\gamma \right) }>0,\\ \check{p}_{2}= & {}\; \frac{(2\beta +6\Gamma -3)\theta -4\gamma (1-\Gamma )}{ 2\left( \theta -2\gamma \right) }=\frac{(6\Gamma -3)\theta +2\left( \beta \theta -2\gamma (1-\Gamma )\right) }{2\left( \theta -2\gamma \right) }>0. \end{aligned}$$

We check for nonnegativity of demands:

$$\begin{aligned} d_{1}^{\ge \eta }=s_{1}\theta +\beta -p_{1}=\frac{1}{4}\left( 2\beta -1\right) >0, \end{aligned}$$

and

$$\begin{aligned} d_{1}^{\eta -1}= & {}\; -\gamma s_{2}-p_{1}+1-\beta \\= & {}\; \frac{1}{4\left( \theta -2\gamma \right) }\left( 7\theta +6\gamma -8\Gamma \left( \theta +2\gamma \right) -6\theta \beta +4\beta \gamma \right) \\> & {}\; 0\Leftrightarrow \Gamma <\frac{7\theta +6\gamma -6\theta \beta +4\beta \gamma }{8\left( \theta +2\gamma \right) }. \end{aligned}$$

If $\Gamma \ge \frac{7\theta +6\gamma -6\theta \beta +4\beta \gamma }{ 8\left( \theta +2\gamma \right) }$, then demand $d_{1}^{\eta -1}$ will be set equal to zero.

Finally,

$$\begin{aligned} d_{2}=\frac{1}{2\left( \theta -2\gamma \right) }\left( \left( 3\theta +2\gamma \right) \left( 2\Gamma -1\right) +2\left( \theta \beta -\gamma \right) \right) >0. \end{aligned}$$

1.4 Proof of Proposition 5

To solve for a Markov-perfect (feedback) Stackelberg equilibrium, we solve the game backward, that is, we start by the second stage.

$$\begin{aligned} d_{1}^{{ \ge \eta }} & = s_{1} \theta + \beta - p_{1} , \\ d_{1}^{{\eta - 1}} & = - \gamma s_{2} - p_{1} + 1 - \beta , \\ d_{2} & = 1 - (d_{1}^{{\eta - 1}} + d_{1}^{{ \ge \eta }} ) + s_{2} \theta - p_{2} , \\ & = \gamma s_{2} + 2p_{1} - \left( {s_{1} - s_{2} } \right)\theta - p_{2} , \\ \end{aligned}$$

1.5 Second-period equilibrium

For any given $s_{2}$ announced by the government, the manufacturer solves the following optimization problem:

$$\begin{aligned} \max _{p_{2}\ge 0}\Pi _{2}=p_{2}\left( \gamma s_{2}+2p_{1}-\left( s_{1}-s_{2}\right) \theta -p_{2}\right) \text {.} \end{aligned}$$

Introduce the manufacturer’s Lagrangian

$$\begin{aligned} {\mathcal {L}}_{M_{2}}\left( p_{2},\lambda _{2}\right) =p_{2}\left( \gamma s_{2}+2p_{1}-\left( s_{1}-s_{2}\right) \theta -p_{2}\right) +\lambda _{2}p_{2}, \end{aligned}$$

where $\lambda _{2}$ is the Lagrange multiplier appended to the constraint $p_{2}\ge 0$. The first-order optimality conditions are

$$\begin{aligned} \frac{\partial {\mathcal {L}}_{M_{2}}}{\partial p_{2}}= & {}\; \gamma s_{2}+2p_{1}-\left( s_{1}-s_{2}\right) \theta -2p_{2}+\lambda _{2}=0, \\ \lambda _{2}\ge & {}\; 0,p_{2}\ge 0,\lambda _{2}p_{2}=0. \end{aligned}$$

Solving the first equation, gives

$$\begin{aligned} p_{2}\left( s_{2}\right) =\frac{\gamma s_{2}+2p_{1}-\left( s_{1}-s_{2}\right) \theta +\lambda _{2}}{2}. \end{aligned}$$

Now, we consider the second-period government’s optimization problem, which is given by

$$\begin{aligned} \min _{s_{2}\ge 0}\ C_{2}=s_{2}.d_{2}. \end{aligned}$$

Substituting for $p_{2}$ in $d_{2}$, the above optimization problem becomes

$$\begin{aligned} \min _{s_{2}\ge 0}\ C_{2}=s_{2}.\left( p_{1}-\frac{1}{2}\left( \lambda _{2}+\theta \left( s_{1}-s_{2}\right) -\gamma s_{2}\right) \right) . \end{aligned}$$

Introduce the second-period government’s Lagrangian

$$\begin{aligned} {\mathcal {L}}_{2}\left( s_{2},\mu _{2}\right) =s_{2}\left( p_{1}-\frac{1}{2} \left( \lambda _{2}+\theta \left( s_{1}-s_{2}\right) -\gamma s_{2}\right) \right) +\eta _{2}s_{2}, \end{aligned}$$

where $\eta _{2}$ is the Lagrange multiplier appended to the constraint $s_{2}\ge 0$. The first-order optimality conditions are

$$\begin{aligned}{} & {} \frac{\partial {\mathcal {L}}_{2}}{\partial s_{2}}=p_{1}-\frac{1}{2} \left( \lambda _{2}+\theta \left( s_{1}-2s_{2}\right) -2\gamma s_{2}\right) +\eta _{2}=0, \\{} & {} \eta _{2}\le 0,s_{2}\ge 0,\ \ \ \eta _{2}s_{2}=0. \end{aligned}$$

Solving the first equation, we obtain

$$\begin{aligned} s_{2}\left( s_{1},p_{1}\right) =\frac{1}{\theta +\gamma }\left( \frac{1}{2} \lambda _{2}-\eta _{2}-p_{1}+\frac{1}{2}\theta s_{1}\right) . \end{aligned}$$

(16)

Substituting in $p_{2}$ yields

$$\begin{aligned} p_{2}\left( s_{1},p_{1}\right) =\frac{1}{4}\left( 2p_{1}+3\lambda _{2}-2\eta _{2}-\theta s_{1}\right) . \end{aligned}$$

(17)

The second-period demand is given by

$$\begin{aligned} d_{2}\left( s_{1},p_{1}\right) =\frac{1}{4}\left( 2p_{1}-2\eta _{2}-\lambda _{2}-\theta s_{1}\right) . \end{aligned}$$

(18)

To wrap up, in (16) and (17), we express the second-period strategies in terms of the first-period decision variables.

1.6 First-period equilibrium (or overall equilibrium problem)

The manufacturer overall optimization problem is as follows:

$$\begin{aligned} \max _{p_{1}\ge 0}\Pi =p_{1}(d_{1}^{\eta -1}+d_{1}^{\ge \eta })+p_{2}\left( s_{1},p_{1}\right) d_{2}\left( s_{1},p_{1}\right) , \end{aligned}$$

(19)

where $p_{2}\left( s_{1},p_{1}\right)$ and $d_{2}\left( s_{1},p_{1}\right)$ have been determined in the previous step and the product $p_{2}\left( s_{1},p_{1}\right) d_{2}\left( s_{1},p_{1}\right)$ plays the role of a salvage value in the current optimization problem.

Substituting for the second-period equilibrium strategies, the above optimization problem becomes:

$$\begin{aligned} \max _{{p_{1} \ge 0}} \Pi = & p_{1} \left( {\frac{{2\left( {\theta + \gamma } \right) - \gamma \lambda _{2} + 2\gamma \eta _{2} + \left( {2\theta + \gamma } \right)\left( {\theta s_{1} - 2p_{1} } \right)}}{{2\left( {\theta + \gamma } \right)}}} \right) \\ & + \frac{1}{{16}}\left( {2p_{1} + 3\lambda _{2} - 2\eta _{2} - \theta s_{1} } \right)\left( {2p_{1} - 2\eta _{2} - \lambda _{2} - \theta s_{1} } \right) \\ \end{aligned}$$

(20)

The Lagrangian is given by

$$\begin{aligned} {\mathcal {L}}_{M}\left( p_{1},\lambda _{1}\right)= & {}\; p_{1}\left( \frac{2\left( \theta +\gamma \right) -\gamma \lambda _{2}+2\gamma \eta _{2}+\left( 2\theta +\gamma \right) \left( \theta s_{1}-2p_{1}\right) }{2\left( \theta +\gamma \right) }\right) \\{} & {} +\frac{1}{16}\left( 2p_{1}+3\lambda _{2}-2\eta _{2}-\theta s_{1}\right) \left( 2p_{1}-2\eta _{2}-\lambda _{2}-\theta s_{1}\right) +\lambda _{1}p_{1}, \end{aligned}$$

where $\lambda _{1}$ is the Lagrange multiplier appended to the constraint $p_{1}\ge 0$.

The first-order optimality conditions give

$$\begin{aligned} p_{1}\left( s_{1}\right)= & {}\; \frac{4\left( \theta +\gamma \right) \left( 1+\lambda _{1}\right) +\left( \lambda _{2}-2\eta _{2}\right) \left( \theta -\gamma \right) +\left( 3\theta +\gamma \right) \theta s_{1}}{14\theta +6\gamma }, \\ p_{1}\ge & {} 0,\lambda _{1}\ge 0,\lambda _{1}p_{1}=0. \end{aligned}$$

Now, we turn to the government’s optimization problem. Substituting for $p_{1}\left( s_{1}\right)$ in (16)–(18) and in the demands, we get

$$\begin{aligned} p_{2}\left( s_{1}\right)= & {}\; \frac{2\left( \theta +\gamma \right) \left( 1+\lambda _{1}\right) +\left( 11\theta +4\gamma \right) \lambda _{2}-2\left( 4\theta +\gamma \right) \eta _{2}-\left( 2\theta +\gamma \right) \theta s_{1} }{14\theta +6\gamma }, \end{aligned}$$

(21)

$$\begin{aligned} d_{2}\left( s_{1}\right)= & {}\; \frac{2\left( \theta +\gamma \right) \left( 1+\lambda _{1}\right) -\left( 3\theta +2\gamma \right) \lambda _{2}-2\left( 4\theta +\gamma \right) \eta _{2}-\left( 2\theta +\gamma \right) \theta s_{1} }{14\theta +6\gamma }, \end{aligned}$$

(22)

$$\begin{aligned} d_{1}^{\ge \eta }\left( s_{1}\right)= & {}\; \frac{-4\left( \theta +\gamma \right) \left( 1+\lambda _{1}\right) +14\theta \beta +6\beta \gamma +\left( 2\eta _{2}-\lambda _{2}\right) \left( \theta -\gamma \right) +\left( 11\theta +5\gamma \right) \theta s_{1}}{14\theta +6\gamma } \end{aligned}$$

(23)

$$\begin{aligned} d_{1}^{{\eta - 1}} \left( {s_{1} } \right) = & - \frac{{\left( {\left( {10\theta + 6\gamma } \right)\left( {\beta - 1} \right) + 4\theta \left( {\beta + \lambda _{1} } \right)} \right)\left( {\theta + \gamma } \right) + \left( {\lambda _{2} - 2\eta _{2} } \right)\left( {\theta ^{2} + 3\gamma ^{2} + 6\theta \gamma } \right)}}{{14\theta ^{2} + 20\theta \gamma + 6\gamma ^{2} }} \\ & - \frac{{\left( {3\theta ^{2} + 3\gamma ^{2} + 8\theta \gamma } \right)\theta s_{1} }}{{14\theta ^{2} + 20\theta \gamma + 6\gamma ^{2} }} \\ \end{aligned}$$

(24)

$$\begin{aligned} \min _{s_{1}\ge 0}&\ C=s_{1}.d_{1}^{\ge \eta }+s_{2}.d_{2}, \nonumber \\ \text {subject to: }&d_{1}^{\ge \eta }+d_{1}^{\eta -1}+d_{2}=\Gamma , \end{aligned}$$

(25)

$$\begin{aligned}&\min _{s_{1}\ge 0}\ C=s_{1}\left( \frac{-4\left( \theta +\gamma \right) \left( 1+\lambda _{1}\right) +14\theta \beta +6\beta \gamma +\left( 2\eta _{2}-\lambda _{2}\right) \left( \theta -\gamma \right) +\left( 11\theta +5\gamma \right) \theta s_{1}}{14\theta +6\gamma }\right) \nonumber \\&+\left( \frac{-2\left( \theta +\gamma \right) \left( 1+\lambda _{1}\right) +\left( 3\theta +2\gamma \right) \left( \lambda _{2}-2\eta _{2}\right) +\left( 2\theta +\gamma \right) \theta s_{1}}{7\theta ^{2}+10\theta \gamma +3\gamma ^{2}}\right) \times \end{aligned}$$

(26)

Inserting for the above values in the following optimization problem

$$\begin{aligned}&\left( \frac{2\left( \theta +\gamma \right) \left( 1+\lambda _{1}\right) -\left( 3\theta +2\gamma \right) \lambda _{2}-2\left( 4\theta +\gamma \right) \eta _{2}-\left( 2\theta +\gamma \right) \theta s_{1}}{14\theta +6\gamma }\right) , \end{aligned}$$

(27)

we get

$$\begin{aligned}&\text {subject to}:\frac{1}{14\theta ^{2}+20\theta \gamma +6\gamma ^{2}} \left( 8\theta ^{2}+4\gamma ^{2}+12\theta \gamma -\left( 6\theta ^{2}+2\gamma ^{2}+8\theta \gamma \right) \lambda _{1}-\right. \end{aligned}$$

(28)

$$\begin{aligned}&\left. \left( 5\theta ^{2}+4\gamma ^{2}-11\theta \gamma \right) \lambda _{2}-2\left( 2\theta ^{2}-\gamma ^{2}-\theta \gamma \right) \eta _{2}+\left( 6\theta ^{2}+\gamma ^{2}+5\theta \gamma \right) \theta s_{1}\right) =\Gamma , \end{aligned}$$

(29)

$$\begin{aligned} s_{1}\left( \lambda _{1},\lambda _{2},\eta _{2}\right)= & {}\; \frac{1}{6\theta ^{3}+\theta \gamma ^{2}+5\theta ^{2}\gamma }\left( -8\theta ^{2}-4\gamma ^{2}-12\theta \gamma +14\Gamma \theta ^{2}+6\Gamma \gamma ^{2}+20\Gamma \theta \gamma \right. \nonumber \\{} & {} \left. +2\left( 3\theta ^{2}+\gamma ^{2}+4\theta \gamma \right) \lambda _{1}+\left( 5\theta ^{2}+4\gamma ^{2}-11\theta \gamma \right) \lambda _{2}\right. \nonumber \\{} & {} \left. +2\left( 2\theta ^{2}-\gamma ^{2}-\theta \gamma \right) \eta _{2}\right) . \end{aligned}$$

(30)

$$\begin{aligned} p_{1}\left( \lambda _{1},\lambda _{2},\eta _{2}\right) =\frac{\left( 14\Gamma \theta ^{2}+6\Gamma \gamma ^{2}+\left( 14\theta ^{2}+6\gamma ^{2}+20\theta \gamma \right) \lambda _{1}+\left( 7\theta ^{2}+3\gamma ^{2}-12\theta \gamma \right) \lambda _{2}+20\Gamma \theta \gamma \right) }{ 28\theta ^{2}+26\theta \gamma +6\gamma ^{2}}. \end{aligned}$$

(31)

The Lagrangian is given by

$$\begin{aligned} {\mathcal {L}}_{1}\left( s_{1},\mu \right)= & {} s_{1}\left( \frac{-4\left( \theta +\gamma \right) \left( 1+\lambda _{1}\right) +14\theta \beta +6\beta \gamma +\left( 2\eta _{2}-\lambda _{2}\right) \left( \theta -\gamma \right) +\left( 11\theta +5\gamma \right) \theta s_{1}}{14\theta +6\gamma }\right) \\{} & {} +\left( \frac{-2\left( \theta +\gamma \right) \left( 1+\lambda _{1}\right) +\left( 3\theta +2\gamma \right) \left( \lambda _{2}-2\eta _{2}\right) +\left( 2\theta +\gamma \right) \theta s_{1}}{7\theta ^{2}+10\theta \gamma +3\gamma ^{2}}\right) \times \\{} & {} \left( \frac{2\left( \theta +\gamma \right) \left( 1+\lambda _{1}\right) -\left( 3\theta +2\gamma \right) \lambda _{2}-2\left( 4\theta +\gamma \right) \eta _{2}-\left( 2\theta +\gamma \right) \theta s_{1}}{14\theta +6\gamma }\right) \\{} & {} +\frac{\mu }{14\theta ^{2}+20\theta \gamma +6\gamma ^{2}}\left( 8\theta ^{2}+4\gamma ^{2}+12\theta \gamma -\left( 6\theta ^{2}+2\gamma ^{2}+8\theta \gamma \right) \lambda _{1}\right. \\{} & {} \left. -\left( 5\theta ^{2}+4\gamma ^{2}-11\theta \gamma \right) \lambda _{2}\right. \\{} & {} \left. -2\left( 2\theta ^{2}-\gamma ^{2}-\theta \gamma \right) \eta _{2}+\left( 6\theta ^{2}+\gamma ^{2}+5\theta \gamma \right) \theta s_{1}\right. \\{} & {} \left. -\left( 14\theta ^{2}+20\theta \gamma +6\gamma ^{2}\right) \Gamma \right) \\{} & {} +\eta _{1}s_{1}, \end{aligned}$$

where $\mu$ and $\eta _{1}$ are the Lagrange multipliers appended to the target constraint and $s_{1}\ge 0$, respectively.

The first-order optimality conditions are

$$\begin{aligned}{} & {} \frac{\partial {\mathcal {L}}_{1}}{\partial s_{1}}=-\frac{1}{14\theta +6\gamma } \left( 4\left( \theta +\gamma \right) \left( 1+\lambda _{1}\right) -14\theta \beta -6\beta \gamma +\left( \lambda _{2}-2\eta _{2}\right) \left( \theta -\gamma \right) \right. \\ \left. -22\theta ^{2}s_{1}-10\theta \gamma s_{1}\right) \\+\frac{\theta \left( 2\theta +\gamma \right) }{\left( 7\theta +3\gamma \right) ^{2}\left( \theta +\gamma \right) }\left( 2\left( 1+\lambda _{1}\right) \left( \theta +\gamma \right) -3\theta \lambda _{2}-2\gamma \lambda _{2}-\theta \eta _{2}\right. \\ \left. +\gamma \eta _{2}-2\theta ^{2}s_{1}-\theta \gamma s_{1}\right) \\ -\theta \mu \frac{6\theta ^{2}+5\theta \gamma +\gamma ^{2}}{14\theta ^{2}+20\theta \gamma +6\gamma ^{2}}\left( 6\theta ^{2}\lambda _{1}-4\gamma ^{2}-12\theta \gamma -8\theta ^{2}+5\theta ^{2}\lambda _{2}+2\gamma ^{2}\lambda _{1}+4\gamma ^{2}\lambda _{2}\right. \\ \left. +8\theta \gamma \lambda _{1}-11\theta \gamma \lambda _{2}\right) \\ +\eta _{1}=0 \\ \\ \frac{\partial {\mathcal {L}}_{1}}{\partial \mu }=\left( 8\theta ^{2}+4\gamma ^{2}+12\theta \gamma -\left( 6\theta ^{2}+2\gamma ^{2}+8\theta \gamma \right) \lambda _{1}\right. \\ \left. -\left( 5\theta ^{2}+4\gamma ^{2}-11\theta \gamma \right) \lambda _{2}\right. \\ \left. -2\left( 2\theta ^{2}-\gamma ^{2}-\theta \gamma \right) \eta _{2}+\left( 6\theta ^{2}+\gamma ^{2}+5\theta \gamma \right) \theta s_{1}\right. \\ \left. -\left( 14\theta ^{2}+20\theta \gamma +6\gamma ^{2}\right) \Gamma \right) =0 \\ \\ \eta _{1}\le 0,s_{1}\ge 0,\ \ \ \eta _{1}s_{1}=0. \end{aligned}$$

From the second condition, we can get $s_{1}$ as function of the model’s parameters and the Lagrange multipliers, that is,

$$\begin{aligned} s_{2}\left( \lambda _{2},\eta _{2}\right)= & {}\; -\frac{\left( 2\theta \gamma -14\theta ^{2}-6\gamma ^{2}\right) \lambda _{2}+\left( 14\theta ^{2}+6\gamma ^{2}+20\theta \gamma \right) \left( 1+\eta _{2}-\Gamma \right) }{21\theta ^{3}+37\theta ^{2}\gamma +19\theta \gamma ^{2}+3\gamma ^{3}}, \end{aligned}$$

(32)

Inserting for $s_{1}$ in $p_{1}\,$we get

$$\begin{aligned} p_{2}\left( \lambda _{2},\eta _{2}\right)= & {}\; \frac{\left( 7\theta ^{2}+3\gamma ^{2}+10\theta \gamma \right) \left( 1-\Gamma \right) +\left( 14\theta +17\gamma \right) \theta \lambda _{2}-\left( 14\theta +6\gamma \right) \theta \eta _{2}}{21\theta ^{2}+16\theta \gamma +3\gamma ^{2}}, \end{aligned}$$

(33)

Substituting for $s_{1}\left( \lambda _{1},\lambda _{2},\eta _{2}\right)$ in $s_{2}\left( s_{1}\right)$ and $p_{2}\left( s_{1}\right)$, we obtain

$$s_{2} = {\text{ }} - \frac{{\left( {14\theta ^{2} + 6\gamma ^{2} + 20\theta \gamma } \right)\left( {1 - \Gamma } \right)}}{{21\theta ^{3} + 37\theta ^{2} \gamma + 19\theta \gamma ^{2} + 3\gamma ^{3} }}\; - \frac{{\left( {14\theta ^{2} + 6\gamma ^{2} + 20\theta \gamma } \right)\left( {1 - \Gamma } \right)}}{{21\theta ^{3} + 37\theta ^{2} \gamma + 19\theta \gamma ^{2} + 3\gamma ^{3} }},{\text{ }}$$

(34)

$$\begin{aligned} p_{2}= & {}\; \frac{\left( 7\theta ^{2}+3\gamma ^{2}+10\theta \gamma \right) \left( 1-\Gamma \right) }{21\theta ^{2}+16\theta \gamma +3\gamma ^{2}}. \end{aligned}$$

(35)

We have four cases to consider:

• 1. $s_{2}>0$ and $p_{2}>0$ $\Rightarrow$ $\eta _{2}=\lambda _{2}=0$. Then,

  $$\begin{aligned} \eta _{2}= & {}\; \Gamma -1<0, \end{aligned}$$

  (36)
  $$\begin{aligned} p_{2}\left( \lambda _{2},\eta _{2}\right)= & {}\; 1-\Gamma >0. \end{aligned}$$

  (37)

  Clearly, $s_{2}$ is negative and therefore a contradiction.

• 2. $s_{2}=0$ and $p_{2}>0$ $\Rightarrow$ $\eta _{2}<0$ and $\lambda _{2}=0$. Then,

  $$\begin{aligned} \eta _{2}= & {}\; \frac{\left( 7\theta ^{2}+3\gamma ^{2}+10\theta \gamma \right) \left( \Gamma -1\right) }{11\theta \gamma }<0, \end{aligned}$$

  (38)
  $$\begin{aligned} \lambda _{2}= & {}\; \frac{\left( 7\theta ^{2}+3\gamma ^{2}+10\theta \gamma \right) \left( \Gamma -1\right) }{11\theta \gamma }>0. \end{aligned}$$

  (39)

• 3. $s_{2}=0$ and $p_{2}=0$ $\Rightarrow$ $\eta _{2}<0$ and $\lambda _{2}>0$. Then,

  $$\begin{aligned} s_{2}= & {}\; \frac{2\left( 7\theta +3\gamma \right) \left( \Gamma -1\right) }{14\theta ^{2}+17\gamma \theta }<0, \end{aligned}$$

  (40)
  $$\begin{aligned} \lambda _{2}= & {}\; \frac{\left( \Gamma -1\right) \left( 7\theta ^{2}+3\gamma ^{2}+10\theta \gamma \right) }{\theta \left( 14\theta +17\gamma \right) }<0, \end{aligned}$$

  (41)

  The fact that $\lambda _{2}$ is strictly negative is a contradiction.

• 4. $s_{2}>0$ and $p_{2}=0$ $\Rightarrow$ $\eta _{2}=0$ and $\lambda _{2}>0$. Then, a contradiction.

Consequently, the only admissible solution is

$$\begin{aligned} s_{2}=0,p_{2}=1-\Gamma ,\eta _{2}=\Gamma -1,\lambda _{2}=0. \end{aligned}$$

Substituting for these values in (32) and (33), we obtain

$$\begin{aligned} s_{1}\left( \lambda _{1}\right)= & {}\; \frac{1}{6\theta ^{3}+\theta \gamma ^{2}+5\theta ^{2}\gamma }\left( -8\theta ^{2}-4\gamma ^{2}-12\theta \gamma +14\Gamma \theta ^{2}+6\Gamma \gamma ^{2}+20\Gamma \theta \gamma \right. \\{}\; & {}\; \left. +2\left( 3\theta ^{2}+\gamma ^{2}+4\theta \gamma \right) \lambda _{1}+2\left( 2\theta ^{2}-\gamma ^{2}-\theta \gamma \right) \left( \Gamma -1\right) \right) ,\\ p_{1}\left( \lambda _{1}\right)= & {}\; \frac{\left( 14\Gamma \theta ^{2}+6\Gamma \gamma ^{2}+\left( 14\theta ^{2}+6\gamma ^{2}+20\theta \gamma \right) \lambda _{1}+20\Gamma \theta \gamma \right) }{28\theta ^{2}+26\theta \gamma +6\gamma ^{2}}. \end{aligned}$$

Clearly, $p_{1}\left( \lambda _{1}\right)$ is strictly positive and therefore $\lambda _{1}$ must be equal to zero. Therefore, the final value of $p_{1}$ is

$$\begin{aligned} p_{1}=\frac{\left( \theta +\gamma \right) \Gamma }{2\theta +\gamma }, \end{aligned}$$

(42)

and

$$\begin{aligned} s_{1}=\frac{6\Gamma \theta +4\Gamma \gamma -4\theta -2\gamma }{\theta \left( 2\theta +\gamma \right) }, \end{aligned}$$

which is positive for

$$\begin{aligned} \Gamma >\frac{2\theta +\gamma }{3\theta +2\gamma }. \end{aligned}$$

To determine the Lagrange multiplier associated with the target constraint, it suffices to substitute for the equilibrium values in $\frac{\partial {\mathcal {L}}_{1}}{\partial s_{1}}=0$ to obtain

$$\begin{aligned} \mu =-\frac{2\theta ^{3}\left( 32\Gamma -23+7\beta \right) +\theta ^{2}\left( 27\beta \gamma -93\gamma +137\Gamma \gamma \right) +\theta \left( 91\Gamma \gamma ^{2}-57\gamma ^{2}\right) +\gamma ^{3}\left( 19\Gamma +3\beta -11\right) +16\theta \beta \gamma ^{2}}{2\theta \left( 3\theta +\gamma \right) \left( \theta +\gamma \right) \left( 2\theta +\gamma \right) ^{3}}. \end{aligned}$$

We check for the nonnegativity of demands. First, we have

$$\begin{aligned} d_{2}=\gamma s_{2}+2p_{1}-\left( s_{1}-s_{2}\right) \theta -p_{2}=1-\Gamma >0. \end{aligned}$$

Using the constraint $d_{1}^{\ge \eta }+d_{1}^{\eta -1}+d_{2}=\Gamma$, we get

$$\begin{aligned} d_{1}^{\ge \eta }+d_{1}^{\eta -1}+1-\Gamma =\Gamma \Leftrightarrow d_{1}^{\ge \eta }+d_{1}^{\eta -1}=2\Gamma -1>0. \end{aligned}$$

So the total demand in the first period is positive. Moreover,

$$\begin{aligned} d_{1}^{\ge \eta }=s_{1}\theta +\beta -p_{1}=\frac{1}{2\theta +\gamma } \left( \Gamma \left( 5+3\gamma \right) +\beta \gamma -4\theta +2\left( \theta \beta -\gamma \right) \right) . \end{aligned}$$

Using $\Gamma >\frac{2\theta +\gamma }{3\theta +2\gamma }$, we have

$$\begin{aligned} d_{1}^{\ge \eta }= & {}\; \frac{1}{2\theta +\gamma }\left( \Gamma \left( 5+3\gamma \right) +\beta \gamma -4\theta +2\left( \theta \beta -\gamma \right) \right) \\> & {}\; \frac{1}{2\theta +\gamma }\left( \frac{2\theta +\gamma }{3\theta +2\gamma }\left( 5+3\gamma \right) +\beta \gamma -4\theta +2\left( \theta \beta -\gamma \right) \right) \\= & {}\; \frac{\theta \left( 3\beta -1\right) +\gamma \left( 2\beta -1\right) +5\left( 1-\theta \right) }{3\theta +2\gamma }>0. \end{aligned}$$

Consider now $d_{1}^{\eta -1}:$

$$\begin{aligned} d_{1}^{\eta -1}= & {}\; -\gamma s_{2}-p_{1}+1-\beta =-p_{1}+1-\beta =1-\frac{ \left( \theta +\gamma \right) \Gamma +\beta \left( 2\theta +\gamma \right) }{2\theta +\gamma } \\> & {}\; 0\Leftrightarrow \frac{\left( \theta +\gamma \right) \Gamma +\beta \left( 2\theta +\gamma \right) }{2\theta +\gamma }<1\Leftrightarrow \Gamma <\frac{ \left( 1-\beta \right) \left( 2\theta +\gamma \right) }{\left( \theta +\gamma \right) }. \end{aligned}$$

If $\Gamma >\frac{\left( 1-\beta \right) \left( 2\theta +\gamma \right) }{ \left( \theta +\gamma \right) }$, then $d_{1}^{\eta -1}=0$.

1.7 Proof of Proposition 6

The manufacturer profits and subsidy costs in both settings are given by:

$$\begin{aligned}{}\; & {}\; \check{C}= \frac{(8\beta ^{2}+(64 \Gamma -36)\beta +96 \Gamma ^{2}-112 \Gamma +32) \theta ^{2}+4( \beta ^2 16\Gamma ^2+\beta ^2-24 \Gamma -3\beta +\dfrac{ 37}{4}) \gamma \theta -8(\beta -\dfrac{1}{2})^{2} \gamma ^{2}}{ 8\theta (2\gamma -\theta )^{2}}\\{}\; & {}\; \check{\Pi }=\frac{(40 \Gamma ^2+(24 \beta -36) \Gamma +4(\beta -\dfrac{3}{2})^2) \theta ^2+16 \gamma (2 \Gamma ^2+(\beta - \dfrac{9}{2}) \Gamma -\beta +\dfrac{3}{2} )\theta +(32(\Gamma ^2-\Gamma +\dfrac{1}{2})) \gamma ^2}{(2 \gamma -\theta )^2}\\{}\; & {}\; {\hat{C}}=\frac{4((\dfrac{3}{2} \Gamma -1) \theta +\gamma (\Gamma -\dfrac{1}{2} ))((2 \beta +5 \Gamma -4) \theta +\gamma (\beta +3 \Gamma -2))}{(2 \theta +\gamma )^2 \theta }\\{}\; & {}\; {\hat{\Pi }}=\frac{(3 \gamma +4 \theta ) \Gamma ^2+(-3 \gamma -5 \theta ) \Gamma +2 \theta +\gamma }{2 \theta +\gamma } \end{aligned}$$

Considering the circumstances in propositions 4 and 5, $\check{C}-{\hat{C}}$ and $\check{\Pi } -{\hat{\Pi }}$ are increasing in $\Gamma$ and positive at $\Gamma =\frac{ 2\theta +\gamma }{3\theta +2\gamma }$, that is to say, $\check{C}>{\hat{C}}$ and $\check{\Pi }>{\hat{\Pi }}$.